`@` `\text {dnammv}`

`a,`

`4x(x^2-x-1)-(x^2-2)(x+3)`

`= 4x^3-4x^2-4x- [x^2(x+3)-2(x+3)]`

`= 4x^3-4x^2-4x- (x^3+3x^2-2x-6)`
`= 4x^3-4x^2-4x-x^3-3x^2+2x+6`

`= 3x^3 - 7x^2-2x+6`

`b,`

`(x+5)(x+7)-7x(x+3)`

`= x(x+7)+5(x+7)-7x^2-21x`

`= x^2+7+5x+35-7x^2-21x`

`= -6x^2-16x+35`

`c,`

`x(x^2-x-2)-(x+5)(x-1)`

`= x^3-x^2-2x- [x(x-1)+5(x-1)]`

`= x^3-x^2-2x- (x^2-x+5x-5)`

`= x^3-x^2-2x - x^2 + x -5x+5`

`= x^3-2x^2- 4x+5`

`d,`

`(x+5)(x+7)-(x-4)(x+3)`

`= x(x+7)+5(x+7)- [x(x+3)-4(x+3)]`

`= x^2+7x+5x+35 - (x^2+3x-4x-12)`

`= x^2+12x+35 - x^2+x+12`

`= 13x+47`

1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)

\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)

\(=\dfrac{1}{2}x^3+x^2-15x-18\)

2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)

\(=4x^3+6x^2-6x^2-9x+10x+15\)

\(=4x^3+x+15\)

3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)

\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)

\(=3x^5-x^4+5x^3+10x^2+26x-5\)

4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)

\(=\left(x^2-1\right)\left(x-2\right)\)

\(=x^3-2x^2-x+2\)

\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)

\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)

\(< =>\left(1-x\right)\left(8x-4\right)=0\)

\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)

\(\left(x-2\right)\left(x+1\right)=x^2-4\)

\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)

\(< =>\left(x-2\right)\left(x+1-x-2\right)=0\)

\(< =>-1\left(x-2\right)=0\)

\(< =>2-x=0< =>x=2\)

\(2x^3+3x^2-32x=48\)

\(< =>x^2\left(2x+3\right)-16\left(2x+3\right)=0\)

\(< =>\left(x^2-16\right)\left(2x+3\right)=0\)

\(< =>\left(x-4\right)\left(x+4\right)\left(2x+3\right)=0\)

\(< =>\hept{\begin{cases}x=4\\x=-4\\x=-\frac{3}{2}\end{cases}}\)

a) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=\left(x^2+3x+1\right)^2+x\)

\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)=\left(x^2+3x+1\right)^2+x\)

\(\Leftrightarrow\left(t-1\right)\left(t+1\right)=t^2+x\) (với \(t=x^2+3x+1\))

\(\Leftrightarrow t^2-1=t^2+x\)

\(\Leftrightarrow x=-1\).

b) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)=\left(x^2+8x+11\right)^2+2x\)

\(\Leftrightarrow\left(x^2+8x+7\right)\left(x^2+8x+15\right)=\left(x^2+8x+11\right)^2+2x\)

\(\Leftrightarrow\left(t-4\right)\left(t+4\right)=t^2+2x\) (với \(t=x^2+8x+11\))

\(\Leftrightarrow t^2-16=t^2+2x\)

\(\Leftrightarrow x=-8\)

c) \(\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)=63\)

\(\Leftrightarrow\left(x^3-1\right)\left(x^3+1\right)=63\)

\(\Leftrightarrow x^6-1=63\)

\(\Leftrightarrow x^6=64\)

\(\Leftrightarrow x=\pm2\)

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\(a,\left(4x+1\right)\left(x-3\right)-\left(x-7\right)\left(4x-1\right)=15\\ \Leftrightarrow4x^2+x-12x-3-\left(4x^2-28x-x+7\right)-15=0\\ \Leftrightarrow4x^2-11x-3-4x^2+29x-7-15=0\\ \Leftrightarrow18x=25\\ \Leftrightarrow x=\dfrac{25}{18}\)

Vậy \(x=\dfrac{25}{18}\)

\(b,\left(x+1\right)\left(x^2-x+1\right)-x\left(x^2-3\right)=4\\ \Leftrightarrow x^3+1-x^3+3x-4=0\\ \Leftrightarrow3x-3=0\\ \Leftrightarrow x=1\)

Vậy \(x=1\)

\(c,\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)-6x=0\\ \Leftrightarrow x^3-27+5x-x^3-6x=0\\ \Leftrightarrow-x-27=0\\ \Leftrightarrow x=-27\)

Vậy \(x=-27\)

\(d,\left(5x-1\right)\left(5x+1\right)=25x^2-7x+15\\ \Leftrightarrow25x^2-1-25x^2+7x-15=0\\ \Leftrightarrow7x-16=0\\ \Leftrightarrow x=\dfrac{16}{7}\)

Vậy \(x=\dfrac{16}{7}\)

\(7,D\\ 8,D\\ 9,B\\ 10,B\\ 11,B\\ 12,C\\ 13,B\\ 14,A\)

b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x^2+1\right)=4\)

\(\Leftrightarrow x^3-1-x^3-x=4\)

\(\Leftrightarrow-x=5\)

hay x=-5

c: Ta có: \(\left(2x-1\right)^3+\left(x+2\right)^3-9x\left(x+1\right)\left(x-1\right)=7\)

\(\Leftrightarrow8x^3-12x^2+6x-1+x^3+6x^2+12x+8-9x^3+9x=7\)

\(\Leftrightarrow-6x^2+27x=0\)

\(\Leftrightarrow-3x\left(2x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{9}{2}\end{matrix}\right.\)

a, \(\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

b, \(3\left(x-2\right)+13⋮x-2\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

c, \(x\left(x+7\right)+2⋮x+7\Rightarrow x+7\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)